Problem: Let $g(x)=x\cos(x)$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\cos(x)+x\sin(x)$ (Choice B) B $\cos(x)-x\sin(x)$ (Choice C) C $-\sin(x)$ (Choice D) D $-\cos(x)$
Solution: $g(x)$ is the product of two, more basic, expressions: $x$ and $\cos(x)$. Therefore, the derivative of $g$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x\cos(x)\right) \\\\ &=\dfrac{d}{dx}\left(x\right)\cos(x)+x\dfrac{d}{dx}(\cos(x))&&\gray{\text{The product rule}} \\\\ &=1\cdot\cos(x)+x\cdot (-\sin(x))&&\gray{\text{Differentiate }x\text{ and }\cos(x)} \\\\ &=\cos(x)-x\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $g'(x)=\cos(x)-x\sin(x)$